How to calculate the two limits? The Next CEO of Stack OverflowCompute $lim limits_xtoinfty (fracx-2x+2)^x$limits of the sequence $n/(n+1)$How to calculate $lim_xto1left(frac1+cos(pi x)tan^2(pi x)right)^!x^2$Calculate the limit of integralHow to evaluate $lim_xtoinftyarctan (4/x)/ |arcsin (-3/x)|$?Is there a way to get at this limit problem algebraically?Calculate $lim_x to infty(x+1)e^-2x$How to calculate $lim_nto infty fracn^nn!^2$?Calculate the limit: $lim limits_n rightarrow infty frac 4(n+3)!-n!n((n+2)!-(n-1)!)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$

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How to calculate the two limits?



The Next CEO of Stack OverflowCompute $lim limits_xtoinfty (fracx-2x+2)^x$limits of the sequence $n/(n+1)$How to calculate $lim_xto1left(frac1+cos(pi x)tan^2(pi x)right)^!x^2$Calculate the limit of integralHow to evaluate $lim_xtoinftyarctan (4/x)/ |arcsin (-3/x)|$?Is there a way to get at this limit problem algebraically?Calculate $lim_x to infty(x+1)e^-2x$How to calculate $lim_nto infty fracn^nn!^2$?Calculate the limit: $lim limits_n rightarrow infty frac 4(n+3)!-n!n((n+2)!-(n-1)!)$How to solve the limit $limlimits_xto infty (x arctan x - fracxpi2)$










3












$begingroup$



I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$




For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.



For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?










share|cite











$endgroup$











  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago















3












$begingroup$



I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$




For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.



For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?










share|cite











$endgroup$











  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago













3












3








3





$begingroup$



I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$




For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.



For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?










share|cite











$endgroup$





I got stuck on two exercises below
$$
limlimits_xrightarrow +infty left(frac2pi arctan x right)^x \
lim_xrightarrow 3^+ fraccos x ln(x-3)ln(e^x-e^3)
$$




For the first one , let $y=(frac2pi arctan x )^x $, so $ln y =xln (frac2pi arctan x )$, the right part is $infty cdot 0$ type, but seemly, the L 'hopital's rule is useless. PS: I know the $infty cdot 0$ can be become to $fracinftyinfty$ or $frac00$. But when I use the L 'hopital's rule to the $fracinftyinfty$ or $frac00$ the calculation is complex and useless.



For the second one , it is $fracinftyinfty$ type, also useless the L 'hopital's rule is. How to calculate it ?







limits






share|cite















share|cite













share|cite




share|cite








edited 1 hour ago







lanse7pty

















asked 2 hours ago









lanse7ptylanse7pty

1,8411823




1,8411823











  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago
















  • $begingroup$
    For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
    $endgroup$
    – Arturo Magidin
    2 hours ago















$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago




$begingroup$
For $0timesinfty$ types, you can algebraically transform them into either $frac00$ or $fracinftyinfty$, where you can try using L'Hopital's (though it may not help): just remember that $ab = fraca frac1b $. And, L'Hopital's Rule is applicable for $fracinftyinfty$ indeterminates...
$endgroup$
– Arturo Magidin
2 hours ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    Without L'Hospital
    $$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$



    Now, by Taylor for large values of $x$
    $$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
    $$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
    $$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
    $$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.






      share|cite|improve this answer









      $endgroup$




















        0












        $begingroup$

        You can solve the first one using



        • $arctan x + operatornamearccotx = fracpi2$

        • $lim_yto 0(1-y)^1/y = e^-1$

        • $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$

        begineqnarray* left(frac2pi arctan x right)^x
        & stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
        & stackrelx to +inftylongrightarrow & e^-frac2pi
        endeqnarray*



        The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider




        • $fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.





        share|cite









        $endgroup$













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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$






          share|cite|improve this answer











          $endgroup$

















            2












            $begingroup$

            Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$






            share|cite|improve this answer











            $endgroup$















              2












              2








              2





              $begingroup$

              Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$






              share|cite|improve this answer











              $endgroup$



              Rewrite $inftycdot 0$ as $infty cdot dfrac1infty$. Now you can apply L'Hopital's rule: $$lim_xto +inftydfracleft(ln 2/picdotarctan x right)1/x=lim_xto +inftydfracpi/2cdot arctan x-1/x^2cdot dfrac11+x^2=-dfracpi 2lim_xto +inftyarctan xcdot dfracx^21+x^2$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 1 hour ago

























              answered 2 hours ago









              Paras KhoslaParas Khosla

              2,736423




              2,736423





















                  1












                  $begingroup$

                  Without L'Hospital
                  $$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$



                  Now, by Taylor for large values of $x$
                  $$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
                  $$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
                  $$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
                  $$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached






                  share|cite|improve this answer









                  $endgroup$

















                    1












                    $begingroup$

                    Without L'Hospital
                    $$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$



                    Now, by Taylor for large values of $x$
                    $$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
                    $$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
                    $$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
                    $$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached






                    share|cite|improve this answer









                    $endgroup$















                      1












                      1








                      1





                      $begingroup$

                      Without L'Hospital
                      $$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$



                      Now, by Taylor for large values of $x$
                      $$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
                      $$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
                      $$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
                      $$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached






                      share|cite|improve this answer









                      $endgroup$



                      Without L'Hospital
                      $$y=left(frac2pi arctan (x) right)^ximplies log(y)=x logleft(frac2pi arctan (x) right) $$



                      Now, by Taylor for large values of $x$
                      $$arctan (x)=fracpi 2-frac1x+frac13 x^3+Oleft(frac1x^4right)$$
                      $$frac2pi arctan (x) =1-frac2pi x+frac23 pi x^3+Oleft(frac1x^4right)$$ Taylor again
                      $$logleft(frac2pi arctan (x) right)= -frac2pi x-frac2pi ^2 x^2+Oleft(frac1x^3right)$$
                      $$log(y)=xlogleft(frac2pi arctan (x) right)= -frac2pi -frac2pi ^2 x+Oleft(frac1x^2right)$$ Just continue with Taylor using $y=e^log(y)$ if you want to see not only the limit but also how it is approached







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      Claude LeiboviciClaude Leibovici

                      125k1158136




                      125k1158136





















                          0












                          $begingroup$

                          I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.






                          share|cite|improve this answer









                          $endgroup$

















                            0












                            $begingroup$

                            I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.






                            share|cite|improve this answer









                            $endgroup$















                              0












                              0








                              0





                              $begingroup$

                              I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.






                              share|cite|improve this answer









                              $endgroup$



                              I believe you can apply L'hopital's rule for an indeterminate form like $fracinftyinfty$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 hours ago









                              AdmuthAdmuth

                              185




                              185





















                                  0












                                  $begingroup$

                                  You can solve the first one using



                                  • $arctan x + operatornamearccotx = fracpi2$

                                  • $lim_yto 0(1-y)^1/y = e^-1$

                                  • $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$

                                  begineqnarray* left(frac2pi arctan x right)^x
                                  & stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
                                  & stackrelx to +inftylongrightarrow & e^-frac2pi
                                  endeqnarray*



                                  The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider




                                  • $fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.





                                  share|cite









                                  $endgroup$

















                                    0












                                    $begingroup$

                                    You can solve the first one using



                                    • $arctan x + operatornamearccotx = fracpi2$

                                    • $lim_yto 0(1-y)^1/y = e^-1$

                                    • $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$

                                    begineqnarray* left(frac2pi arctan x right)^x
                                    & stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
                                    & stackrelx to +inftylongrightarrow & e^-frac2pi
                                    endeqnarray*



                                    The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider




                                    • $fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.





                                    share|cite









                                    $endgroup$















                                      0












                                      0








                                      0





                                      $begingroup$

                                      You can solve the first one using



                                      • $arctan x + operatornamearccotx = fracpi2$

                                      • $lim_yto 0(1-y)^1/y = e^-1$

                                      • $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$

                                      begineqnarray* left(frac2pi arctan x right)^x
                                      & stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
                                      & stackrelx to +inftylongrightarrow & e^-frac2pi
                                      endeqnarray*



                                      The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider




                                      • $fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.





                                      share|cite









                                      $endgroup$



                                      You can solve the first one using



                                      • $arctan x + operatornamearccotx = fracpi2$

                                      • $lim_yto 0(1-y)^1/y = e^-1$

                                      • $xoperatornamearccotx stackrelstackrelx =cot uuto 0^+= cot ucdot u = cos ucdot fracusin u stackrelu to 0^+longrightarrow 1$

                                      begineqnarray* left(frac2pi arctan x right)^x
                                      & stackrelarctan x = fracpi2-operatornamearccotx= & left( underbraceleft(1- frac2pioperatornamearccotxright)^fracpi2operatornamearccotx_stackrelx to +inftylongrightarrow e^-1 right)^frac2piunderbracexoperatornamearccotx_stackrelx to +inftylongrightarrow 1 \
                                      & stackrelx to +inftylongrightarrow & e^-frac2pi
                                      endeqnarray*



                                      The second limit is quite straight forward as $lim_xto 3+cos x = cos 3$. Just consider




                                      • $fracln(x-3)ln(e^x-e^3)$ and apply L'Hospital.






                                      share|cite












                                      share|cite



                                      share|cite










                                      answered 6 mins ago









                                      trancelocationtrancelocation

                                      13.4k1827




                                      13.4k1827



























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                                          Torre de la Isleta Índice Véase también Referencias Bibliografía Enlaces externos Menú de navegación38°25′58″N 0°23′02″O / 38.43277778, -0.3838888938°25′58″N 0°23′02″O / 38.43277778, -0.38388889Torre de la Illeta de l’Horta o Torre Saleta. Base de datos de bienes inmuebles. Patrimonio Cultural. Secretaría de Estado de CulturaFicha BIC Torre de la Illeta de l’Horta. Dirección General de Patrimonio Cultural. Generalitat ValencianaLugares de interés. Ayuntamiento del CampelloTorre de la Isleta en CastillosNet.org